Injective Modules: Preparatory Material for the Snowbird Summer School on Commutative Algebra

INJECTIVE MODULES: PREPARATORY MATERIAL FOR THE SNOWBIRD SUMMER SCHOOL ON COMMUTATIVE ALGEBRA These notes are intended to give the reader an idea what injective modules are, where they show up, and, to a small extent, what one can do with them. Let R be a commutative Noetherian ring with an identity element. An R- module E is injective if HomR(−, E) is an exact functor. The main messages of these notes are • Every R-module M has an injective hull or injective envelope, denoted by ER(M), which is an injective module containing M, and has the property that any injective module containing M contains an isomorphic copy of ER(M). • A nonzero injective module is indecomposable if it is not the direct sum of nonzero injective modules. Every injective R-module is a direct sum of indecomposable injective R-modules. • Indecomposable injective R-modules are in bijective correspondence with the prime ideals of R; in fact every indecomposable injective R-module is isomorphic to an injective hull ER(R/p), for some prime ideal p of R. • The number of isomorphic copies of ER(R/p) occurring in any direct sum decomposition of a given injective module into indecomposable injectives is independent of the decomposition. • Let (R, m) be a complete local ring and E = ER(R/m) be the injective hull of the residue field of R. The functor (−)∨ = HomR(−, E) has the following properties, known as Matlis duality: (1) If M is an R-module which is Noetherian or Artinian, then M∨∨ M. ∼= (2) If M is Noetherian, then M∨ is Artinian. (3) If M is Artinian, then M∨ is Noetherian. Any unexplained terminology or notation can be found in [1] or [3]. Matlis’ theory of injective modules was developed in the paper [4], and may also be found in [3, § 18] and [2, § 3]. These notes owe a great deal of intellectual debt to Mel Hochster, whose lecture notes are very popular with the organizers of this workshop. However, the organizers claim intellectual property of all errors here. 1. Injective Modules Throughout, R is a commutative ring with an identity element 1 ∈ R. All R-modules M are assumed to be unitary, i.e., 1 · m = m for all m ∈ M. 1<ul style="display: flex;"><li style="flex:1">2</li><li style="flex:1">INJECTIVE MODULES </li></ul>Definition 1.1. An R-module E is injective if for all R-module homomorphisms ϕ : M −→ N and ψ : M −→ E where ϕ is injective, there exists an R-linear homomorphism θ : N −→ E such that θ ◦ ϕ = ψ. Exercise 1.2. Show that E is an injective R-module E if and only if HomR(−, E) is an exact functor, i.e., applying HomR(−, E) takes short exact sequences to short exact sequences. Theorem 1.3 (Baer’s Criterion). An R-module E is injective if and only if every R-module homomorphism a −→ E, where a is an ideal, extends to a homomorphism R −→ E. Proof. One direction is obvious. For the other, if M ⊆ N are R-modules and ϕ : M −→ E, we need to show that ϕ extends to a homomorphism N −→ E. By Zorn’s lemma, there is a module N0 with M ⊆ N0 ⊆ N, which is maximal with respect to the property that ϕ extends to a homomorphism ϕ0 : N0 −→ E. If N0 = N, take an element n ∈ N \N0 and consider the ideal ϕ0 na = N0 :R n. By hypothesis, the composite homomorphism a −→ N0 −→ E extends to a homomorphism ψ : R −→ E. Define ϕ00 : N0 + Rn −→ E by ϕ00(n0 + rn) = ϕ0(n0) + ψ(r). This contradicts the maximality of ϕ0, so we must have N0 = N. ꢀExercise 1.4. Let R be an integral domain. An R-module M is divisible if rM = M for every nonzero element r ∈ R. (1) Prove that an injective R-module is divisible. (2) If R is a principal ideal domain, prove that an R-module is divisible if and only if it is injective. (3) Conclude that Q/Z is an injective Z-module. (4) Prove that any nonzero Abelian group has a nonzero homomorphism to Q/Z. (5) If (−)∨ = HomZ(−, Q/Z) and M is any Z-module, prove that the natural map M −→ M∨∨ is injective. Exercise 1.5. Let R be an A-algebra. (1) Use the adjointness of ⊗ and Hom to prove that if E is an injective A-module, and F is a flat R-module, then HomA(F, E) is an injective R-module. (2) Prove that every R-module can be embedded in an injective R- module. Hint: If M is the R-module, take a free R-module F with a surjection F −→→ HomZ(M, Q/Z). Apply (−)∨ = HomZ(−, Q/Z). Proposition 1.6. Let M = 0 and N be R-modules, and let θ : M ,→ N be a monomorphism. Then the following are equivalent: (1) Every nonzero submodule of N has a nonzero intersection with θ(M). (2) Every nonzero element of N has a nonzero multiple in θ(M). (3) If ϕ ◦ θ is injective for a homomorphism ϕ : N −→ Q, then ϕ is injective. <ul style="display: flex;"><li style="flex:1">INJECTIVE MODULES </li><li style="flex:1">3</li></ul>Proof. (1) =⇒ (2) If n is a nonzero element of N, then the cyclic module Rn has a nonzero intersection with θ(M). (2) =⇒ (3) If not, then ker ϕ has a nonzero intersection with θ(M), contradicting the assumption that ϕ ◦ θ is injective. (3) =⇒ (1) Let N0 be a nonzero submodule of N, and consider the canonical surjection ϕ : N −→ N/N0. Then ϕ is not injective, hence the composition ϕ ◦ θ : M −→ N/N0 is not injective, i.e., N0 contains a nonzero element of θ(M). ꢀDefinition 1.7. If θ : M ,→ N satisfies the equivalent conditions of the previous proposition, we say that N is an essential extension of M. Example 1.8. If R is a domain and Frac(R) is its field of fractions, then R ⊆ Frac(R) is an essential extension. More generally, if S ⊆ R is the set of nonzerodivisors in R, then S−1R is an essential extension of R. Example 1.9. Let (R, m) be a local ring and N be an R-module such that every element of N is killed by a power of m. The socle of N is the submodule soc(N) = 0 :N m. Then soc(N) ⊆ N is an essential extension: if n ∈ N is a nonzero element, let t be the smallest integer such that mtn = 0. Then mt−1n ⊆ soc(N), and mt−1n contains a nonzero multiple of n. Exercise 1.10. Let I be an index set. Then Mi ⊆ Ni is essential for all i ∈ I if and only if ⊕i∈IMi ⊆ ⊕i∈INi is essential. Example 1.11. Let R = C[[x]] which is a local ring with maximal ideal (x), and take N = Rx/R. Every element of N is killed by a power of the maximal ideal, and soc(N) is the 1-dimensional C-vector space generated by [1/x], i.e., the image of 1/x in N. By Example 1.9, soc(N) ⊆ N is an essential <ul style="display: flex;"><li style="flex:1">Q</li><li style="flex:1">Q</li></ul>extension. However N soc(N) ⊆ N N is not an essential extension since the element Y<ul style="display: flex;"><li style="flex:1">ꢀ</li><li style="flex:1">ꢁ</li></ul>[1/x], [1/x2], [1/x3], . . . ∈ NNQdoes not have a nonzero multiple in N soc(N). (Prove!) Proposition 1.12. Let L, M, N be nonzero R-modules. (1) M ⊆ M is an essential extension. (2) Suppose L ⊆ M ⊆ N. Then L ⊆ N is an essential extension if and only if both L ⊆ M and M ⊆ N are essential extensions. (3) Suppose M ⊆ N and M ⊆ Ni ⊆ N with N = ∪iNi. Then M ⊆ N is an essential extension if and only if M ⊆ Ni is an essential extension for every i. (4) Suppose M ⊆ N. Then there exists a module N0 with M ⊆ N0 ⊆ N, which is maximal with respect to the property that M ⊆ N0 is an essential extension. Proof. The assertions (1), (2), and (3) elementary. For (4), let F = {N0 | M ⊆ N0 ⊆ N and N0 is an essential extension of M}. <ul style="display: flex;"><li style="flex:1">4</li><li style="flex:1">INJECTIVE MODULES </li></ul>Then M ∈ F so F is nonempty. If N10 ⊆ N20 ⊆ N30 ⊆ . . . is a chain in F, then ∪iNi0 ∈ F is an upper bound. By Zorn’s Lemma, the set F has maximal elements. ꢀDefinition 1.13. The module N0 in Proposition 1.12 (4) is a maximal essential extension of M in N. If M ⊆ N is essential and N has no proper essential extensions, we say that N is a maximal essential extension of M. Proposition 1.14. Let M be an R-module. The following conditions are equivalent: (1) M is injective; (2) M is a direct summand of every module containing it; (3) M has no proper essential extensions. Proof. (1) =⇒ (2) =⇒ (3) is left as an exercise, and we prove the implication (3) =⇒ (2). Consider an embedding M ,→ E where E is injective. By Zorn’s lemma, there exists a submodule N ⊆ E which is maximal with respect to the property that N ∩ M = 0. This implies that M ,→ E/N is an essential extension, and hence that it is an isomorphism. But then E = M + N so E = M ⊕ N. Since M is a direct summand of an injective module, it must be injective. ꢀProposition 1.15. Let M and E be R-modules. (1) If E is injective and M ⊆ E, then any maximal essential extension of M in E is an injective module, hence is a direct summand of E. (2) Any two maximal essential extensions of M are isomorphic. Proof. (1) Let E0 be a maximal essential extension of M in E and let E0 ⊆ Q be an essential extension. Since E is injective, the identity map E0 −→ E lifts to a homomorphism ϕ : Q −→ E. Since Q is an essential extension of E0, it follows that ϕ must be injective. This gives us M ⊆ E0 ⊆ Q ,→ E, and the maximality of E0 implies that Q = E0. Hence E0 has no proper essential extensions, and so it is an injective module by Proposition 1.14. (2) Let M ⊆ E and M ⊆ E0 be maximal essential extensions of M. Then E0 is injective, so M ⊆ E0 extends to a homomorphism ϕ : E −→ E0. The inclusion M ⊆ E is an essential extension, so ϕ is injective. But then ϕ(E) is an injective module, and hence a direct summand of E0. Since M ⊆ ϕ(E) ⊆ E0 is an essential extension, we must have ϕ(E) = E0. ꢀDefinition 1.16. The injective hull or injective envelope of an R-module M is a maximal essential extension of M, and is denoted by ER(M). Definition 1.17. Let M be an R-module. A minimal injective resolution of M is a complex 0 −→ E0 −→ E1 −→ E2 −→ . . . such that E0 = ER(M), E1 = ER(E0/M), and Ei+1 = ER(Ei/ image(Ei−1)) for all i ≥ 2. <ul style="display: flex;"><li style="flex:1">INJECTIVE MODULES </li><li style="flex:1">5</li></ul>Note that the modules Ei are injective, and that image(Ei) ⊆ Ei+1 is an essential extension for all i ≥ 0. 2. Injectives over a Noetherian Ring Proposition 2.1 (Bass). A ring R is Noetherian if and only if every direct sum of injective R-modules is injective. Proof. We first show that if M is a finitely generated R-module, then ∼Hom (M, ⊕ N ) ⊕ Hom (M, N ). =<ul style="display: flex;"><li style="flex:1">R</li><li style="flex:1">i</li><li style="flex:1">i</li><li style="flex:1">i</li><li style="flex:1">R</li><li style="flex:1">i</li></ul>Independent of the finite generation of M, there is a natural injective homomorphism ϕ : ⊕i HomR(M, Ni) −→ HomR(M, ⊕iNi). If M is finitely generated, the image of a homomorphism from M to ⊕iNi is contained in the direct sum of finitely many Ni. Since Hom commutes with forming finite direct sums, ϕ is surjective as well. Let R be a Noetherian ring, and Ei be injective R-modules. Then for an ideal a of R, the natural map HomR(R, Ei) −→ HomR(a, Ei) is surjective. Since a is finitely generated, the above isomorphism implies that HomR(R, ⊕Ei) −→ HomR(a, ⊕Ei) is surjective as well. Baer’s criterion now implies that ⊕Ei is injective. If R is not Noetherian, it contains a strictly ascending chain of ideals a1 ( a2 ( a3 ( . . . . Let a = ∪iai. The natural maps a ,→ R −→→ R/ai ,→ ER(R/ai) give us Qa homomorphism a −→ i ER(R/ai). The image lies in the submodule ⊕iER(R/ai), (check!) so we have a homomorphism ϕ : a −→ ⊕ ER(R/ai). Lastly, check that ϕ does not extend to homomorphism R −→ ⊕iiER(R/ai). ꢀTheorem 2.2. Let E be an injective module over a Noetherian ring R. Then ∼E ⊕ E (R/p ), =<ul style="display: flex;"><li style="flex:1">i</li><li style="flex:1">R</li><li style="flex:1">i</li></ul>where pi are prime ideals of R. Moreover, any such direct sum is an injective R-module. Proof. The last statement follows from Proposition 2.1. Let E be an injective R-module. By Zorn’s Lemma, there exists a maximal family {Ei} of injective ∼submodules of E such that Ei E (R/p ), and their sum in E is a direct =<ul style="display: flex;"><li style="flex:1">R</li><li style="flex:1">i</li></ul>sum. Let E0 = ⊕Ei, which is an injective module, and hence is a direct summand of E. There exists an R-module E00 such that E = E0 ⊕ E00. If E00 = 0, pick a nonzero element x ∈ E00. Let p be an associated prime of Rx. Then R/p ,→ Rx ⊆ E00, so there is a copy of ER(R/p) contained in E00 and E00 = ER(R/p) ⊕ E000, contradicting the maximality of family {Ei}. ꢀDefinition 2.3. Let a be an ideal of a ring R, and M be an R-module. We say M is a-torsion if every element of M is killed by some power of a. <ul style="display: flex;"><li style="flex:1">6</li><li style="flex:1">INJECTIVE MODULES </li></ul>Theorem 2.4. Let p be a prime ideal of a Noetherian ring R, and let E = ER(R/p) and κ = Rp/pRp, which is the fraction field of R/p. Then x(1) if x ∈ R \ p, then E −→ E is an isomorphism, and so E = Ep; (2) 0 :E p = κ; (3) κ ⊆ E is an essential extension of Rp-modules and E = ER (κ); p(4) E is p-torsion and Ass(E) = {p}; (5) HomR (κ, E) = κ and HomR (κ, ER(R/q)p) = 0 for primes q = p. <ul style="display: flex;"><li style="flex:1">p</li><li style="flex:1">p</li></ul>Proof. (1) κ is an essential extension of R/p by Example 1.8, so E contains a copy of κ and we may assume R/p ⊆ κ ⊆ E. Multiplication by x ∈ R\p is injective on κ, and hence also on its essential extension E. The submodule xE is injective, so it is a direct summand of E. But κ ⊆ xE ⊆ E are essential extensions, so xE = E. (2) 0 :E p = 0 :E pRp is a vector space over the field κ, and hence the inclusion κ ⊆ 0 :E p splits. But κ ⊆ 0 :E p ⊆ E is an essential extension, so 0 :E p = κ. (3) The containment κ ⊆ E is an essential extension of R-modules, hence also of Rp-modules. Suppose E ⊆ M is an essential extension of Rp-modules, pick m ∈ M. Then m has a nonzero multiple (r/s)m ∈ E, where s ∈ R \ p. But then rm is a nonzero multiple of m in E, so E ⊆ M is an essential extension of R-modules, and therefore M = E. (4) Let q ∈ Ass(E). Then there exists x ∈ E such that Rx ⊆ E and 0 :R x = q. Since R/p ⊆ E is essential, x has a nonzero multiple y in R/p. But then the annihilator of y is p, so q = p and Ass(E) = {p}. If a is the annihilator of a nonzero element of E, then p is the only associated prime of R/a ,→ E, so E is p-torsion. (5) For the first assertion, ∼HomR (κ, E) = HomR (R /pR , E) 0 : E = κ. =<ul style="display: flex;"><li style="flex:1">p</li><li style="flex:1">p</li></ul>pRp <ul style="display: flex;"><li style="flex:1">p</li><li style="flex:1">p</li></ul>Since elements of R\q act invertibly on ER(R/q), we see that ER(R/q)p = 0 if q * p. In the case q ⊆ p, we have ∼<ul style="display: flex;"><li style="flex:1">HomR (κ, E (R/q) ) 0 : </li><li style="flex:1">ER(R/q)p = 0 :pR ER(R/q). </li></ul>=RppRp <ul style="display: flex;"><li style="flex:1">p</li><li style="flex:1">p</li></ul>If this is nonzero, then there is a nonzero element of ER(R/q) killed by p, which forces q = p since Ass ER(R/q) = {q}. ꢀWe are now able to strengthen Theorem 2.2 to obtain the following structure theorem for injective modules over Noetherian rings. Theorem 2.5. Let E be an injective over a Noetherian ring R. Then M<ul style="display: flex;"><li style="flex:1">E = </li><li style="flex:1">ER(R/p)αp , </li></ul>p∈Spec R and the numbers αp are independent of the direct sum decomposition. <ul style="display: flex;"><li style="flex:1">INJECTIVE MODULES </li><li style="flex:1">7</li></ul>Proof. Theorem 2.2 implies that a direct sum decomposition exists. By Theorem 2.4 (5), αp is the dimension of the Rp/pRp-vector space HomR (Rp/pRp, Ep),

Injective Modules: Preparatory Material for the Snowbird Summer School on Commutative Algebra

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